Solve ∫ (from 3 to 6) of 3(x^2+1) wrt x

$$\int _ { 3 } ^ { 6 } 3 ( x ^ { 2 } + 1 ) d x$$

Evaluate

$$198$$

Solution Steps

Use the distributive property to multiply 3 by x^{2}+1.

$$\int _{3}^{6}3x^{2}+3\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 3x^{2}+3\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int 3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x+\int 3\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 3 times \frac{x^{3}}{3}.

$$x^{3}+\int 3\mathrm{d}x$$

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{3}+3x$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$6^{3}+3\times 6-\left(3^{3}+3\times 3\right)$$

Simplify.

$$198$$